Using rational roots theorem write the function in factored form p(x)=x^4+6*x^3-13*x^2-66*x+72

P(x) = x^4+6*x^3-13*x^2-66*x+72 ; By trial find out for what value of x, P(x) = 0
P(1) = 0 ==> P(x) = 0 has a root x = 1 ==> x - 1 is a factor of P(x)
Now P(x) = x^4 - x^3 + 7x^3 - 7x^2 - 6x^2 + 6x - 72x + 72
= (x^4 - x^3) + (7x^3 - 7x^2) - (6x^2 - 6x) - (72x - 72)
= x^3 (x - 1) + 7x^2(x - 1) - 6x(x - 1) - 72(x - 1)
= (x - 1){ x^3 + 7x^2 - 6x - 72} = (x - 1) * Q(x) [ Writing Q(x) for the 2nd factor]
By Trial find out dor what value of x, Q(x) = 0. It is found that when x = 3, Q(x) = 0
i.e. Q(3) = 0 . By the same arguement x - 3 is a factor of Q(x0
Now P(x) = (x - 1) {x^3 - 3x^2 + 10x^2 - 30x + 24x - 72}
= (x - 1){(x^3 - 3x^2) + (10x^2 - 30x) + (24x - 72)}
= (x - 1){x^2(x - 3) + 10x(x - 3) + 24(x - 3)}
= (x - 1)(x - 3){ x^2 + 10x + 24} = (x - 1)(x - 3) * R(x) [ writing R(x) for the 3rd factor]
By trial find out for what value od x, R(x) = 0. It is found that when x = - 4, R(x) = 0
i.e. R( - 4) = 0. By the same arguement x + 4 is a factor of R(x) = 0
Now P(x) = (x - 1)(x - 3){x^2 + 10x + 24}
= (x - 1)(x - 3){(x^2 + 4x + 6x + 24} = (x - 1)(x - 3){(x^2 + 4x) +( 6x + 24)}
= (x - 1)(x - 3){x(x + 4) + 6(x + 4)} = (x - 1)(x - 3)(x +4)(x + 6)
}

The rational roots theorem tells you to check for roots at ± (1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72).
In fact there are roots at x = -6, -4, 1, and 3.
Use the factor theorem to convert these roots into factors:
(x + 6) (x + 4) (x - 1) (x - 3)